$\begin{array}{l}
\sin x\cos x\cos 2x\cos 4x\cos 8x = \frac{1}{{16}}\sin 12x\\
\Leftrightarrow 16\sin x\cos x\cos 2x\cos 4x\cos 8x = \sin 12x\\
\Leftrightarrow 8\sin 2x\cos 2x\cos 4x\cos 8x = \sin 12x\\
\Leftrightarrow 4\sin 4x\cos 4x\cos 8x = \sin 12x\\
\Leftrightarrow 2\sin 8x\cos 8x = \sin 12x\\
\Leftrightarrow \sin 16x = \sin 12x\\
\Leftrightarrow \left[ \begin{array}{l}
16x = 12x + k2\pi \\
16x = \pi - 12x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = k2\pi \\
28x = \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{2}\\
x = \frac{\pi }{{28}} + \frac{{k\pi }}{{14}}
\end{array} \right.\\
TH1:x = \frac{{k\pi }}{2}\\
x \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow - \frac{\pi }{2} \le \frac{{k\pi }}{2} \le \frac{\pi }{2} \Leftrightarrow - 1 \le k \le 1 \Rightarrow k \in \left\{ { - 1;0;1} \right\}\\
\Rightarrow x = - \frac{\pi }{2};0;\frac{\pi }{2}\\
TH2:x = \frac{\pi }{{28}} + \frac{{k\pi }}{{14}}\\
x \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right] \Rightarrow - \frac{\pi }{2} \le \frac{\pi }{{28}} + \frac{{k\pi }}{{14}} \le \frac{\pi }{2} \Leftrightarrow - \frac{{15}}{2} \le k \le \frac{{13}}{2} \Rightarrow k \in \left\{ { - 7; - 6;...;6;7} \right\}\\
x = - \frac{{13\pi }}{{28}}; - \frac{{11\pi }}{{28}}; - \frac{{9\pi }}{{28}}; - \frac{{7\pi }}{{28}}; - \frac{{5\pi }}{{28}}; - \frac{{3\pi }}{{28}}; - \frac{\pi }{{28}};\frac{\pi }{{28}};\frac{{3\pi }}{{28}};\frac{{5\pi }}{{28}};\frac{{7\pi }}{{28}};\frac{{9\pi }}{{28}};\frac{{11\pi }}{{28}};\frac{{13\pi }}{{28}}
\end{array}$
